Jack drops a stone from rest off of the top of a bridge that is 24.4 m above the ground. After the stone falls 6.6 m, Jill throws a second stone straight down. Both rocks hit the water at the exact same time. What was the initial velocity of Jill's rock? Assume upward is the positive direction and downward is negative. (Indicate the direction with the sign of your answer.)

Question

Steve · Accepted Answer

dropped stone hits water:
24.4-4.9t^2=0
t=2.23

stone has fallen 6.6m:
4.9t^2 = 6.6
t = 1.16

So, Jill's thrown stone only has 2.23-1.16=1.07 seconds to hit the water:

24+1.07v-4.9*1.07^2 = 0
v = -17.19 m/s

bobpursley · Answer

Jack: d=1/2 g t^2 not time when his stone is 6.6, is t1=sqrt (6.6/4.9) compute that Jill d=v*(t-t1) +1/2 g (t-t1)^2 Now you know d=24.4 solve for Jack time t then, go to Jill equation, solve for v.